Geometry Problem

Click the figure below to see the complete problem 494 about Circular Sector, 90 Degrees, Semicircle, Chord, Parallel.

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Complete Problem 494

Level: High School, SAT Prep, College geometry

## Saturday, August 7, 2010

### Problem 494: Circular Sector, 90 Degrees, Semicircle, Chord, Parallel

Labels:
90,
chord,
circular sector,
degree,
parallel,
semicircle

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Line AB intersect circle O’ at M. M is midpoint of AB . Triangle AO’M is isosceles right triangle

ReplyDeleteQuadrilateral ACBD is a parallelogram and diagonals CD and AB intersect at midpoint M of AB and CD.

E is midpoint of AC (corresponding points of dilation transformation centered at A with dilation factor =1/2 ).

Angle ADC=45 (Angle face 90 degrees arc AM)

Consider circle O’ and point C . We have CE.CA=CM.CD .

Since M and E are the midpoints of CD and CA so CA=CD and CM=CE

triangle ACD is isosceles right triangle.

Triangles ACM congruence with DCE ( case SAS)

So x=DE=AM =O’A*SQRT(2)=2*SQRT(2)

Just confirming Peter's excellent solution by analytical geometry: Let O be (0,0). So A:(0,4),B:(4,0). The two circles are: x^2+y^2=16 & x^2+(y-2)^2=4. Now let C be (c,d) and D be (a,b). The given data can be translated into following eqns: a^+(b-2)^2=4, c^2+d^2=16,bc=(a-4)(d-4) and c^2+(d-4)^2=(a-4)^2+b^2 which when solved give us: a=8/5,b=4/5,c=12/5 and d=16/5.

ReplyDeleteNow we can write the eqn. of AC as y=-x/3+4 and determine E as (6/5,18/5) as the intersection of the smaller circle and AC. Now x^2=(8/5-6/5)^2+(4/5-18/5)^2 or x^2=8 -->x=2√2

Ajit

ang AOC = ang AO'E (isoc tr & A common)

ReplyDeleteang COB = ang DO'O ( OD bisect)

=> AO'E + DO'O = 90° => EO'D = 90°

=> x² = 2² + 2²

x = 2√2

To c.t.o.e.:

ReplyDeleteWill u pl explain what u mean by "ang COB=ang DO'O (OD bisect)". Unfortunately, it isn't very clear how u arrive at EO'D = 90° which, of course, has to be correct.

Vihaan

To Vihaan:

ReplyDeleteOD perpendicular to AD and CD (AD // CD)

BOC isoceles

=> OD bisector

DO'O = arc OD

BOD = 1/2 arc OD ( OB tg to O' and OD chord )

BOC = 2 BOD = arc OD

=> DO'O = BOC

( nice solution of 495 )

Thanks. BTW, I'm very curious to know what "c.t.e.o" stands for!

ReplyDeletevihaanup@gmail.com